I need help babnwnwnnw

A car rental agency charges Shaes Family $25.00 plus .10 per mile that the car has driven. Shae spends $45.00 on the rental car.

S_A_V [24]

She drove 200 miles.

6 0

3 years ago

50 POINTS! WILL MARK BRAINLEST! NEED IN 10 - 15 MINUTES!

masha68 [24]

Answer:

21

(2,14)

14/2=7

14+7=21

21 is final answer

Erika had a 95% chance of guessing wrong

20 can be seen as 100%

the % will go down by 5's because of the original number

.5*20=

This will give you a 95% success rate.

7 0

3 years ago

Calculus piecewise function.

Kipish [7]

Part A

The notation $\lim_{x \to 2^{+}}f(x)$ means that we're approaching x = 2 from the right hand side (aka positive side). This is known as a right hand limit.

So we could start at say x = 2.5 and get closer to 2 by getting to x = 2.4 then to x = 2.3 then 2.2, 2.1, 2.01, 2.001, etc

We don't actually arrive at x = 2 itself. We simply move closer and closer.

Since we're on the positive or right hand side of 2, this means we go with the rule involving x > 2

Therefore f(x) = (x/2) + 1

Plug in x = 2 to find that...

f(x) = (x/2) + 1

f(2) = (2/2) + 1

f(2) = 2

This shows $\lim_{x \to 2^{+}}f(x) = 2$

Then for the left hand limit $\lim_{x \to 2^{-}}f(x)$ , we'll involve x < 2 and we go for the first piece. So,

f(x) = 3-x

f(2) = 3-2

f(2) = 1

Therefore, $\lim_{x \to 2^{-}}f(x) = 1$

===============================================================

Part B

Because $\lim_{x \to 2^{+}}f(x) \ne \lim_{x \to 2^{-}}f(x)$ this means that the limit $\lim_{x \to 2}f(x)$ does not exist.

If you are a visual learner, check out the graph below of the piecewise function. Notice the gap or disconnect at x = 2. This can be thought of as two roads that are disconnected. There's no way for a car to go from one road to the other. Because of this disconnect, the limit doesn't exist at x = 2.

===============================================================

Part C

You'll follow the same type of steps shown in part A.

However, keep in mind that x = 4 is above x = 2, so we'll deal with x > 2 only.

So you'd only involve the second piece f(x) = (x/2) + 1

You should find that f(4) = 3, and that both left and right hand limits equal this value. The left and right hand limits approach the same y value. The limit does exist here. There are no gaps to worry about when x = 4.

===============================================================

Part D

As mentioned earlier, since $\lim_{x \to 4^{+}}f(x) = \lim_{x \to 4^{-}}f(x) = 3$ , this means the limit $\lim_{x \to 4}f(x)$ does exist and it's equal to 3.

As x gets closer and closer to 4, the y values are approaching 3. This applies to both directions.

4 0

2 years ago

On a very cold morning, it was -7°F. As the day went on the temperature rose 3

Nina [5.8K]

AnswgameI ok er: my question about the hidden object

Step-by-step explanation:

Jsbskehvejjhh ha okay

3 0

4 years ago

Read 2 more answers

A jewelry box has a length that is 2 inches longer than the width and a height that is 1 inch smaller than the width. the volume

maxonik [38]

Given that a <span>jewelry box has a length that is 2 inches longer than the width and a height that is 1 inch smaller than the width.

Let the width of the box be w, then the length of the box is given by l = w + 2 and the height of the box is given by h = w - 1.

The volume of a box is given by

$V=l\times w\times h \\ \\ =w(w+2)(w-1) \\ \\ =w(w^2+w-2)=w^3+w^2-2w$

Given that the volume of the box is 140 cubic inches. Thus

$w^3+w^2-2w=140$

Thus, the width of the jewelry box is obtained as follows

$w^3+w^2-2w=140 \\ \\ \Rightarrow w^3+w^2-2w-140=0 \\ \\ \Rightarrow(w-5)(w^2+6w+28)=0 \\ \\ \Rightarrow w=5$

Therefore, the width of the jewelry box is 5 inches.

*Note that $w^2+6w+28=0$ gives complex roots, so we did not use it.
</span>

7 0

3 years ago